Programme d'approximation racine carrée Java avec boucle while

Question

J'essaie de créer un programme qui approximera la racine carrée d'un nombre via le théorème de compression. Si un nombre est supérieur à un (l'exemple simple est 100), il définit l'intervalle sur [1, 100], prend le point central et si le carré de ce nombre est supérieur à 100, il met à jour haut et imprime le nouvel intervalle. Si elle est inférieure, elle se met à jour bas jusqu'à ce que l'espace entre eux soit dans EPSILON. Mon problème est que je n'arrive pas à l'obtenir dans l'instruction else pour mettre à jour bas. Toute la sortie avant cela est correcte. Il imprime les cinq premières lignes correctement puis se termine sans mise à jour faible.

public static void main(String[] args) { 

    System.out.println("Enter a number to find its square root -> "); 

    Scanner cin = new Scanner(System.in); 
    double number = cin.nextDouble(); 
    double low = 0; 
    double high = 0; 
    double midPoint; 
    double midPointSqr; 
    final double EPSILON = .000001; 

     if (number < 0) 
      System.out.printf("The square root of %f is NaN%n", number); 
     else if (number < 1) { 
      low = 0; high = 1; 
     } 
     else { 
      low = 1; 
      high = number; 
      midPoint = (low+high)/2; 
      midPointSqr = (midPoint*midPoint); 
      System.out.printf("[%.6f, %.6f]%n", low, high); 
      high = midPoint; 
      System.out.printf("[%.6f, %.6f]%n", low, high); 
      while ((Math.abs(midPoint*midPoint)-number) >= EPSILON) { 
       midPoint = (low+high)/2; 
       if (midPointSqr > number) { 
        high = midPoint; 
        System.out.printf("[%.6f, %.6f]%n", low, high); 
        midPoint = (low+high)/2; 
        midPointSqr = midPoint*midPoint; 
       } 
       else if (midPointSqr == number) 
        System.out.println("null"); 
       else { 
        low = midPoint; 
        System.out.printf("[%.6f, %.6f]%n", low, high); 
        midPoint = (low+high)/2; 
        midPointSqr = midPoint*midPoint; 
       } 
      } 

     } 
/*Output should look like this 
1 Enter a number to find its square root -> 100 
2 [1.000000 , 100.000000] 
3 [1.000000 , 50.500000] 
4 [1.000000 , 25.750000] 
5 [1.000000 , 13.375000] 
6 [7.187500 , 13.375000] 
7 [7.187500 , 10.281250] 
8 [8.734375 , 10.281250] 
9 [9.507813 , 10.281250] 
10 [9.894531 , 10.281250] 
11 [9.894531 , 10.087891] 
12 [9.991211 , 10.087891] 
13 [9.991211 , 10.039551] 
14 [9.991211 , 10.015381] 
15 [9.991211 , 10.003296] 
16 [9.997253 , 10.003296] 
17 [9.997253 , 10.000275] 
18 [9.998764 , 10.000275] 
19 [9.999519 , 10.000275] 
20 [9.999897 , 10.000275] 
21 [9.999897 , 10.000086] 
22 [9.999991 , 10.000086] 
23 [9.999991 , 10.000039] 
24 [9.999991 , 10.000015] 
25 [9.999991 , 10.000003] 
26 [9.999997 , 10.000003] 
27 [9.999997 , 10.000000] 
28 [9.999999 , 10.000000] 
29 [10.000000 , 10.000000] 
*/ 

Answer 1

Voici ma version de nettoyé votre code:

public static double approximateSqrt(double number) { 
    double low = 0; 
    double high = 0; 
    double midPoint; 
    double midPointSqr; 
    final double EPSILON = .000001; 

    if (number < 0) { 
     System.out.printf("The square root of %f is NaN%n", number); 
     return Double.NaN; 
    } else if (number < 1) { 
     low = 0; 
     high = 1; 
    } else { 
     low = 1; 
     high = number; 
    } 
    midPoint = (low + high)/2; 
    midPointSqr = (midPoint * midPoint); 

    while (Math.abs(midPointSqr - number) >= EPSILON) { 
     System.out.printf("[%.6f, %.6f]%n", low, high); 

     if (midPointSqr > number) { 
      high = midPoint; 
     } else { 
      low = midPoint; 
     } 

     midPoint = (low + high)/2; 
     midPointSqr = (midPoint * midPoint); 
    } 
    System.out.printf("Final. midpoint=%.6f num=%.6f%n", midPoint, number); 
    return midPoint; 
} 

public static void main(String[] args) { 
    System.out.println("Enter a number to find its square root -> "); 
    Scanner cin = new Scanner(System.in); 
    double number = cin.nextDouble(); 
    approximateSqrt(number); 
}